--- title: "Operators and expected energy in a well" subject: "Electrical Engineering Material" chapter: 1 key: "semester_3/electrical_engineering_material:operators-expected-energy-well" source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/operators-expected-energy-well site: BCT Engineering --- # Operators and expected energy in a well Electrical Engineering Material · Chapter 1: Theory of Metals - Asked in: 2076 Chaitra, 4+4 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/operators-expected-energy-well - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-1 ## Question Explain the significance of operators in quantum mechanics. How do you calculate the expected energy value of a particle represented by ψ(x,t) confined at a boundary of 0 to L? ## Short answer Operators represent observables and yield measurable values when acting on the wave function. The expected energy of a particle in an infinite well (0 to L) is ⟨E⟩ = ∫₀ᴸ ψ* (-ħ²/2m)(d²/dx²) ψ dx, giving E_n = n²h²/(8mL²). ## Answer #### Significance of operators In quantum mechanics every measurable quantity (observable) is represented by an **operator**, a mathematical instruction that acts on the wave function $\psi$. | Observable | Operator | |---|---| | Position $x$ | $\hat{x} = x$ | | Momentum $p_x$ | $\hat{p}_x = -j\hbar\dfrac{\partial}{\partial x}$ | | Total energy $E$ | $\hat{E} = j\hbar\dfrac{\partial}{\partial t}$ | | Hamiltonian (KE + PE) | $\hat{H} = -\dfrac{\hbar^2}{2m}\dfrac{\partial^2}{\partial x^2} + V(x)$ | Why they matter: - They give a rule for getting physical information from $\psi$; $\psi$ alone is not measurable. - **Eigenvalue equation:** if $\hat{A}\psi = a\psi$, a measurement of $A$ always gives the exact value $a$. The Schrödinger equation itself is $\hat{H}\psi = E\psi$, so allowed energies are eigenvalues of $\hat{H}$. - **Expectation value:** when $\psi$ is not an eigenfunction, operators give the average result of many measurements: $\langle A\rangle = \int\psi^*\hat{A}\psi\,dx$. - Non-commuting operators (e.g. $\hat{x}$ and $\hat{p}$) lead to the uncertainty principle. #### Expected energy of a particle in a well (0 to L) For an infinite well of width $L$ ($V = 0$ inside), the normalized wave function is $$ \psi_n(x,t) = \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)e^{-jE_nt/\hbar} $$ The expected (average) energy is $$ \langle E\rangle = \int_0^L \psi^*\,\hat{E}\,\psi\,dx = \int_0^L \psi^*\left(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\right)\psi\,dx $$ Using $\dfrac{\partial^2\psi}{\partial x^2} = -\left(\dfrac{n\pi}{L}\right)^2\psi$: $$ \begin{aligned} \langle E\rangle &= \frac{\hbar^2}{2m}\left(\frac{n\pi}{L}\right)^2\int_0^L\psi^*\psi\,dx \\ &= \frac{\hbar^2}{2m}\left(\frac{n\pi}{L}\right)^2\cdot\frac{2}{L}\int_0^L\sin^2\frac{n\pi x}{L}\,dx \\ &= \frac{\hbar^2n^2\pi^2}{2mL^2}\cdot\frac{2}{L}\cdot\frac{L}{2} = \frac{n^2h^2}{8mL^2} \end{aligned} $$ (The same result follows from $\hat{E} = j\hbar\,\partial/\partial t$, since $j\hbar\,\partial\psi/\partial t = E_n\psi$.) **Result:** $\langle E\rangle = \dfrac{n^2h^2}{8mL^2}$. Because $\psi_n$ is an eigenfunction of the energy operator, every measurement gives this same value; there is no spread in energy. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In quantum mechanics, an observable is represented by: A. A number B. An operator acting on the wave function C. The square of psi D. A classical variable Answer: B. Operators extract physical information from psi. (Key point: Operators) 2. If A psi = a psi, a measurement of A gives: A. Zero B. An average of several values C. A random value D. Exactly a every time Answer: D. For an eigenfunction, the measurement result is the eigenvalue with no spread. (Key point: Eigenvalue equation) 3. The expectation value of A for normalised psi is: A. Integral of psi* A psi dx B. Integral of A psi dx C. Integral of |psi|^2 dx D. A applied to psi at x = 0 Answer: A. = integral psi* A psi dx. (Key point: Expectation value) 4. The expected energy for the infinite well of width L in state n is: A. n h^2/8mL^2 B. h^2/8mL^2n^2 C. n^2 h^2/8mL^2 D. n^2 h^2/2mL^2 Answer: C. psi_n is an eigenfunction of H, giving E_n = n^2 h^2/(8mL^2). 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